#include <iostream>
#include <cstdio>
#include <cmath>
using namespace std;

const int MAXN = 1e7 + 10;

inline int read()
{
  char c = getchar();
  int x = 0, f = 1;
  while (c < '0' || c > '9')
  {
    if (c == '-')
      f = -1;
    c = getchar();
  }
  while (c >= '0' && c <= '9')
  {
    x = x * 10 + c - '0';
    c = getchar();
  }
  return x * f;
}
const double Pi = acos(-1.0);
struct complex
{
  double x, y;
  complex(double xx = 0, double yy = 0) { x = xx, y = yy; }
} a[MAXN], b[MAXN];
complex operator+(complex a, complex b) { return complex(a.x + b.x, a.y + b.y); }
complex operator-(complex a, complex b) { return complex(a.x - b.x, a.y - b.y); }
complex operator*(complex a, complex b) { return complex(a.x * b.x - a.y * b.y, a.x * b.y + a.y * b.x); } //不懂的看复数的运算那部分
int N, M;
int l, r[MAXN];
int limit = 1;
void fast_fast_tle(complex *A, int type)
{
  for (int i = 0; i < limit; i++)
    if (i < r[i])
      swap(A[i], A[r[i]]);                  //求出要迭代的序列
  for (int mid = 1; mid < limit; mid <<= 1) //待合并区间的长度的一半
  {
    complex Wn(cos(Pi / mid), type * sin(Pi / mid)); //单位根
    for (int R = mid << 1, j = 0; j < limit; j += R) //R是区间的长度，j表示前已经到哪个位置了
    {
      complex w(1, 0);                          //幂
      for (int k = 0; k < mid; k++, w = w * Wn) //枚举左半部分
      {
        complex x = A[j + k], y = w * A[j + mid + k]; //蝴蝶效应
        A[j + k] = x + y;
        A[j + mid + k] = x - y;
      }
    }
  }
}
int main()
{
    int N = read(), M = read();
    for (int i = 0; i <= N; i++)
      a[i].x = read();
    for (int i = 0; i <= M; i++)
      b[i].x = read();
    while (limit <= N + M)
      limit <<= 1, l++;
    for (int i = 0; i < limit; i++)
      r[i] = (r[i >> 1] >> 1) | ((i & 1) << (l - 1));
    // 在原序列中 i 与 i/2 的关系是 ： i可以看做是i/2的二进制上的每一位左移一位得来
    // 那么在反转后的数组中就需要右移一位，同时特殊处理一下奇数
    fast_fast_tle(a, 1);
    fast_fast_tle(b, 1);
    for (int i = 0; i <= limit; i++)
      a[i] = a[i] * b[i];
    fast_fast_tle(a, -1);
    for (int i = 0; i <= N + M; i++)
      printf("%d ", (int)(a[i].x / limit + 0.5));
  }
    
  return 0;
}